Question: Teresa is maintaining a camp fire. She can keep the fire burning for $4$ hours with $6$ logs. She wants to know how many logs $(y)$ she needs to keep the fire burning for $18$ hours. She assumes all logs are the same. How many logs does Teresa need to maintain the fire for $18$ hours?
We're dealing with a proportional relationship, so each ratio of logs to hours must be equivalent. Here's one way to write the proportion: $\dfrac{y\text{ logs}}{18\text{ hours}} = \dfrac{6\text{ logs}}{4\text{ hours}}$ Now, solve the proportion for $y$ : $\begin{aligned} \dfrac{y}{18} &= \dfrac{6}{4} \\\\ \dfrac{y}{18} &= \dfrac{3}{2} \\\\ y &= \dfrac{3}{2} \cdot 18 \\\\ y &= \dfrac{3\cdot 18}{2} \\\\ y &= \dfrac{3\cdot \stackrel{9}{\cancel{18}} }{\underset{1}{\cancel2}} \\\\ y &= \dfrac{3 \cdot 9}{1} \\\\ y &= \dfrac{27}{1} \\\\ y &= 27 \end{aligned}$ Teresa needs $27$ logs to maintain the fire for $18$ hours.